Aug
16
2011
Simple Interest problem!! pls help!!!!?
In one year, Charlotte earned the same interest from an investment at 8% annual interest as an investment at 12% annual interest. She had invested 1500$ more in the 8% account. How much money did she have invested at 12%?
anyone who is willing to give a detailed procedure that even a 1st grader can understand please do so. Im in a really confused state and my algebra teacher isnt really making it any easier, can anyone help pls?^_^
This conversation was published at Accounting Services of Charlotte, NC
David GAugust 16, 2011 at 5:43 pm
x is the amount of $ invested that is in common. Interest is on the total amount of $ therefore the side with the 8% interest will look like the following with the $1500 extra: (x+1500) whereas the side of the 12% will just have the x. The following equation is the set up: 8%(x+$1500)=12%(x). Change the % to decimal form by moving the decimal point 2 places to the left and removing the percent sign. It will look like this: . 08(x+$1500)= . 12x Now distribute the . 08 through to get . 08x+ (. 08 times $1500) = . 12x Now multiply what is inside parenthesis . 08x +120= . 12x Now subtract . 08x from both sides to combine the x. 120=. 12x-. 08x Now combine the x terms 120=. 04x In order to get 1x we need to multiply both sides by 25 because 25 times . 04=1 (Think of needing 25 four cent pieces if they existed to make $1. ) When we do this we get 3000=x Do a check by replacing 3000 in for x to see if both sides are equal. It does checks, therefore 3000+1500 is invested at 8% and 3000 is invested at 12%. Hope it helps. If not let me know. . .